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3 years ago

What element forms an ion with an electronic configuration of [kr] and a –2 charge?

2 answers:

8 0

Must be Selenium (Se). It is two space away from Kr, which means that it needs two extra electron to be like a noble gas, Kr.

nekit [7.7K]3 years ago

6 0

Must be Selenium (Se). It is two space away from Kr, which means that it needs two extra electron to be like a noble gas, Kr.

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What environment will produce the largest crystal sizes?

VMariaS [17]

The pressure should be low as low pressure can melt the crystals easier way
According to Gay lussac's law pressure is directly proportional to temperature so temperature should be also low as it's general any thing need to be cooled down inorder to get crystallised
Now for amount of dissolved solids the solution should be super saturated like magma .It denotes that the particles present as solute must be greater than the solubility.
Cooling time also should be long inorder to give time to crystallization

3 0

2 years ago

The solubility of nitrogen gas at 25 ◦C and 1 atm is 6.8×10−4 mol/L. If the partial pressure of nitrogen gas in air is 0.76 atm,

iris [78.8K]

Answer:

Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L

Explanation:

More the pressure of the gas, more will be its solubility.

So, for two different pressure, the relation between them is shown below as:-

$\frac {P_1}{C_1}=\frac {P_2}{C_2}$

Given ,

P₁ = 1 atm

P₂ = 0.76 atm

C₁ = 6.8 × 10⁻⁴ mol/L

C₂ = ?

Using above equation as:

$\frac{1\ atm}{6.8\times 10^{-4}\ mol/L}=\frac{0.76\ atm}{C_2}$

$C_2=\frac{0.76\times 6.8\times 10^{-4}}{1}\ mol/L$

<u>Concentration of dissolved nitrogen = 5.2 × 10⁻⁴ mol/L</u>

5 0

3 years ago

In some places timber companies remove all the trees from entire hillside when they are harvesting logs and farmers till the soi

laila [671]

Answer:

The timber companies removing all the trees from entire hillside when they are harvesting logs is a practice that could cause the following when it is time to plant in spring:

1. It could affect the quality of plant needed nutrients in the soil and beneficial microorganism population in the soil which could impact negatively the planting season.

2. Trees serves as protection of soil nutrients against wind erosions, so the soil nutrients would be affected.

The farmers tilling the soil means preparing the soil for a farming season as the following effect:

1. Helps control weed for the planting season.

2. This practice could further encourage soil erosion if not done well.

3. Tilling the soil could make leftover plants from the felling of trees mix well with the soil and also add nutrients to the soil when they decay.

8 0

2 years ago

Aluminium +hydrochloride acid ________+_________

REY [17]

Answer:

Aluminium Chloride + Hydrogen

Please vote for Brainliest and I hope this helps!

4 0

2 years ago

For the reaction 2N2O5(g) <---> 4NO2(g) + O2(g), the following data were colected:

KonstantinChe [14]

Answer:

a) The reaction is first order, that is, order 1. Option C is correct.

b) The half life of the reaction is 23 minutes. Option B is correct

c) The initial rate of production of NO2 for this reaction is approximately = (3.7 × 10⁻⁴) M/min. Option has been cut off.

Explanation:

First of, we try to obtain the order of the reaction from the data provided.

t (minutes) [N2O5] (mol/L)

0 1.24x10-2

10 0.92x10-2

20 0.68x10-2

30 0.50x10-2

40 0.37x10-2

50 0.28x10-2

70 0.15x10-2

Using a trial and error mode, we try to obtain the order of the reaction. But let's define some terms.

C₀ = Initial concentration of the reactant

C = concentration of the reactant at any time.

k = rate constant

t = time since the reaction started

T(1/2) = half life

We Start from the first guess of zero order.

For a zero order reaction, the general equation is

C₀ - C = kt

k = (C₀ - C)/t

If the reaction is indeed a zero order reaction, the value of k we will obtain will be the same all through the set of data provided.

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = (0.0124 - 0.0092)/10 = 0.00032 M/min

At t = 20 minutes, C = 0.0068 M

k = (0.0124 - 0.0068)/20 = 0.00028 M/min

At t = 30 minutes, C = 0.0050 M

k = (0.0124 - 0.005)/30 = 0.00024 M/min

It's evident the value of k isn't the same for the first 3 trials, hence, the reaction isn't a zero order reaction.

We try first order next, for first order reaction

In (C₀/C) = kt

k = [In (C₀/C)]/t

C₀ = 0.0124 M

At t = 10 minutes, C = 0.0092 M

k = [In (0.0124/0.0092)]/10 = 0.0298 /min

At t = 20 minutes, C = 0.0068 M

k = 0.030 /min

At t = 30 minutes, C = 0.0050 M

k = 0.0303

At t = 40 minutes

k = 0.0302 /min

At t = 50 minutes,

k = 0.0298 /min

At t = 60 minutes,

k = 0.031 /min

This shows that the reaction is indeed first order because all the answers obtained hover around the same value.

The rate constant to be taken will be the average of them all.

Average k = 0.0302 /min.

b) The half life of a first order reaction is related to the rate constant through this relation

T(1/2) = (In 2)/k

T(1/2) = (In 2)/0.0302

T(1/2) = 22.95 minutes = 23 minutes.

c) The initial rate of production of the product at the start of the reaction

Rate = kC (first order)

At the start of the reaction C = C₀ = 0.0124M and k = 0.0302 /min

Rate = 0.0302 × 0.0124 = 0.000374 M/min = (3.74 × 10⁻⁴) M/min

3 0

3 years ago